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Organic chemistry problem

  1. Nov 13, 2006 #1
    • OP warned about not using the homework template
    Say we have a mixture of 10 mL of methane and propane. For a complete combustion, we need 41 mL of O2. Calculate the composition of the mixture in terms of volume, if atmospheric temperature and pressure are constants.

    The equations are:
    CH4 + O2 --> CO2 + H2O
    C3H8 + O2 --> CO2 + H2O

    I do not understand the question: am I asked to calculate a percentage of volumes, e.g. VCH4/VC3H8, or simply to find the volumes of CH4 and C3H8 separately (if so, how?)? Am I considering the reactives or the products of the reaction? :confused:

    These calculations must come in somewhere in the solution:
    I know that all gazes have the same molecular volume at normal temperature and pressure: 22.4L/mol. Therefore I can determine the quantity of moles of each component. In the initial mixture (before reaction), there is:
    0.224mol of CH4 + C3H8
    0.918mol of O2.

    Thx. for helping!
     
    Last edited: Nov 13, 2006
  2. jcsd
  3. Nov 10, 2016 #2

    Bystander

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    You want the volume fraction of the initial mixture, (methane/(methane + propane)), or (propane/(methane + propane)).
    This is correct, given the ideal gas constraint.
    Balance these two equations.
    Solve the system of two equations in two unknowns, number of moles ∝ 10 and number of moles ∝ f(41).
     
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